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In greater detail, is it possible to tile space such the number of cells that a ray of a length $l$ will pass through (or, equivalently, the ratio of cell number to $l$) tends to become uniform in all directions as $l \rightarrow \infty$, assuming (if it makes a difference) that $l$ starts in the center of some cell?

For one dimensional space, this is trivial; if one marks of regularly-spaced grid points along a line, a ray of a given length starting in the center of one grid section will pass through the same number of grid sections whether it points in either of the two available directions; equivalently, the ratio of cell number to length will approach the same value for rays of arbitrary length in either direction.

In 2 and 3 dimensions, though, a trivial square or cubic grid does not have the same property; rays of equal length aligned perpendicular to edges or faces will pass through different numbers of cells than rays that pass through corners, and the ratios of cell numbers to lengths for rays of arbitrary length will approach different values in each direction. It is not obvious how one could construct a grid that did not have that property. So, is it in fact possible to do? Or are all grids in dimensions higher than 1 necessarily anisotropic?

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    If you tile a one-dimensional space, an infinite ray in either direction will pass through an infinite number of cells. A "ray" of finite length (i.e., a line segment) passes through a finite number of cells, but as we let its length grow without bound, the number of cells grows without bound. I don't consider that something that "tends towards a constant value." I think perhaps you wanted the ratio of cells to the length of the segment, which _does_ tend toward a constant in one dimension. – 2017-01-30
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    No, I mean that at sufficient large lengths, the number is close to the same in any direction. Perhaps there is a clearer way to phrase that, but the ratio of cells to length doesn't seem like an improvement; in any specific direction, it will tend towards a constant value even on a square grid, but it will still vary with direction on a square grid. – 2017-01-31
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    The way we know a grid is anisotropic is precisely _because_ the ratio of cells to length converges to different values in different directions, for example as we can see happens in a square grid. In order to converge to different values, it must first converge in each direction! The number of cells touched does not converge. – 2017-01-31
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    It seems that we are talking past each other. I don't disagree with you, though I may have done a bad job of properly expressing myself; my original conception of the problem just multiplies everything by a factor of $l$, which merely scales the values for each direction, and does not effect their equivalence. I have attempted to edit the question to make it clearer, approaching the problem from both points of view. – 2017-01-31
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    Draw points from a [Poisson process](http://www.mathematik.uni-ulm.de/stochastik/fundl/paper/frey/survey2/node2.html) and define the grid to be their [Voronoi diagram](https://en.wikipedia.org/wiki/Voronoi_diagram)...? – 2017-01-31
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    Let me try to put this another way: $\lim_{x\to\infty} \frac{\lfloor x\rfloor}{x}=1,$ but $\lim_{x\to\infty} \lfloor x\rfloor=\infty.$ Since $\lim_{x\to\infty} \frac{\lfloor 2x\rfloor}{x}=2,$ we can see that $\lfloor 2x\rfloor$ "grows faster" than $\lfloor x\rfloor$ (in the sense defined by these two limits); but $\lim_{x\to\infty} \lfloor 2x\rfloor=\infty,$ the same limit as $\lim_{x\to\infty} \lfloor x\rfloor.$ So no, number of cells is not at all "equivalent" to ratio of number of cells to length. – 2017-01-31
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    Yes, I understand. Let *me* try to put it another way: let $c(l,v)$ be the cell count for a ray of length $l$ in direction $v$. Then $\forall v.\lim \limits_{x \to \infty}c(x,v) = \infty$, but $\lim \limits_{x \to \infty}c(x,v)/c(x,u)$ may be a finite value. Isotropy is indicated by $\exists C. \forall v.\lim \limits_{x \to \infty}c(x,v)/x = C$, but this is equivalent to $\forall v.\forall u.\lim \limits_{x \to \infty}\frac{c(x,v)/x}{c(x,u)/x} = 1 \equiv \forall v.\forall u.\lim \limits_{x \to \infty}\frac{c(x,v)}{c(x,u)} = 1 \equiv \forall v.\forall u. c(x,v) \approx c(x,u)\ where\ x >> 0$ – 2017-02-01
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    @Rahul That seems reasonable, but is there a proof? If so, it raises the further question: are there *regular* isotropic grids in dimensions greater than 1? – 2017-02-01
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    The segment in one dimension must be viewed as a ball of radius $l$. Otherwise, I think it isn't possible that such lattice exists in euclidian space with respect to the usual metric. – 2017-09-02

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